Solve the Quadratic Inequality: When is x² + 8x - 9 Less Than Zero?

To solve for the values of $x$ where $y = x^2 + 8x - 9$ is less than zero, we will follow these steps:

• Find the roots of the quadratic equation $x^2 + 8x - 9 = 0$ using the quadratic formula.
• Determine the intervals on the x-axis formed by these roots.
• Test values from each interval in the inequality to determine where the function is negative.

First, we calculate the roots of $x^2 + 8x - 9 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 8$, $c = -9$.

The discriminant is calculated as:

$b^2 - 4ac = 8^2 - 4(1)(-9) = 64 + 36 = 100$

Since the discriminant is positive, there are two distinct real roots.

The roots are:

$x = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2}$

This gives us roots $x = 1$ and $x = -9$.

Now, we analyze the sign of $y = x^2 + 8x - 9$ around these root intervals:

• For $x < -9$, choose a number like $x = -10$.
• For $-9 < x < 1$, choose a number like $x = 0$.
• For $x > 1$, choose a number like $x = 2$.

Substituting these test points into the function:

• For $x = -10$: $y = (-10)^2 + 8(-10) - 9 = 100 - 80 - 9 = 11 > 0$
• For $x = 0$: $y = (0)^2 + 8(0) - 9 = -9 < 0$
• For $x = 2$: $y = (2)^2 + 8(2) - 9 = 4 + 16 - 9 = 11 > 0$

Therefore, the function $y = x^2 + 8x - 9$ is negative in the interval $-9 < x < 1$.

Considering the inequality $f(x) < 0$, we conclude:

The solution to the problem is $-9 < x < 1$, aligning with answer choice 4.